A vertical `U`-tube has a liquid up to a height `h`. If the tube is slowly rotated to an angular speed `omega=sqrt(g/h)` find the (i) heights of the liquid column, ii pressures at the points `1, 2` and `3` in the limbs in steady state.

Answer

`(rhogh)/4`

Explanation

`(dp)/(dx)=rho(omega^(2)x)impliesdp=rhoomega^(2)x dx`
integrating both sides, we get

`int_(p_(1))^(p_(3))dp=rhoomega^(2)int_(-h)^(2h)x dx`
`p_(3)-p_(1)=rhoomega^2[(x^(2))/2]=(rhoomega^(2))/2[4h^(2)-h^(2)]`
`rho(h_(3)-h_(1))=(3rhoomega^(2)h^(2))/2=(3rho)/2h^(2)(g/h)`
`h_(3)=h_(1)=(3h)/2`..........i
But `h_(3)+h_(1)=2h`.........ii
After solving `h_(3)=(7h)/4` and `h_(1)=h/4`
`p_(3)=p_(0)+rog((7h)/4)` and `p_(1)=p_(0)+rhog(h/4)`
and `p_(3)-p_(2)=(rhoomega^(2))/2[x^(2)]_(0)^(2)=(rhoomega^(2))/2(rh)^(2)`
`p_(3)-p_(2)=rho/2(g/h).4h^(2)=2rhogh`
`p_(0)+rhog((7h)/4)-p_(2)=2rhoghimpliesp_(2)=p_(0)-(rhogh)/4`